Question

Electromagnetic radiation is emitted by accelerating charges. The rate at which energy is emitted from an accelerating charge that has charge q and acceleration a is given by dE dt = q2a2 6pP0c3 where c is the speed of light. (a) Verify that this equation is dimensionally correct. (b) If a proton with a kinetic energy of 6.0 MeV is traveling in a particle accelerator in a circular orbit of radius 0.750 m, what fraction of its energy does it radiate per second? (c) Consider an electron orbiting with the same speed and radius. What fraction of its energy does it radiate per second?

Answer 1

Answer:

a)  $\frac{dE}{dt}=\frac{q^2a^2}{4\pi E_0C^3}$

b)  $x=\frac{6.0}{4.3*10^{-32}}$

c) $x'=\frac{6.0}{2.22*10^{-3}}$

Explanation:

From the question we are told that:

Kinetic energy of Proton$K.E_p= 6.0 MeV$

Radius $r=0.750$

Energy $\frac{dE}{dt}=\frac{q^2a^2}{4\pi E_0C^3}$

a)

$\frac{dE}{dt}=\frac{q^2a^2}{4\pi E_0C^3}$

b)

Generally the equation for acceleration of proton is mathematically given by

$a_p=\frac{v^2}{r}$

Where

Speed of Proton particle is

$V_p=2.12 *10^3 m/s$

$a_p=\frac{(2.12 *10^3)^2}{0.750}\\a_p=1.27*10^{10}m/s^2$

Therefore

$\frac{dE}{dt}=\frac{q^2a^2}{4\pi E_0C^3}$

$\frac{dE}{dt}=\frac{9*10^9*(1.6*10^{-19}^2*(1.27*10^{10})^2)}{(3*10^8)^3}$

$\frac{dE}{dt}=6.881810^{-51}J/s$

Energy radiated per sec

$mev=\frac{6.881810^{-51}}{1.6*10^{-19}}\\mev=4.3*10^{-32}ev$

Therefore the Fraction of its energy  it radiates per second is given as

$x=\frac{ K.E_p}{mev}$

$x=\frac{6.0}{4.3*10^{-32}}$

c)

Generally the equation for acceleration of proton is mathematically given by

$a_p=\frac{v^2}{r}$

Where

Speed of Proton particle is

$V_p=2.5 *10^7 m/s$

$a_p=\frac{(2.5 *10^7)^2}{0.750}\\a_p=0.33*10^{14}m/s^2$

Therefore

$\frac{dE}{dt}=\frac{q^2a^2}{4\pi E_0C^3}$

$\frac{dE}{dt}=\frac{9*10^9*(1.6*10^{-19}^2*(0.33*10^{14})^2)}{(3*10^8)^3}$

$\frac{dE}{dt}=59.26^{-25}J/s$

Energy radiated per sec

$mev=\frac{59.26^{-25}}{1.6*10^{-19}}\\mev=2.22*10^{-3}ev$

Therefore the Fraction of its energy  it radiates per second is given as

$x'=\frac{ K.E_p}{mev}$

$x'=\frac{6.0}{2.22*10^{-3}}$